That is a regular octogan (8 sided shape)
The sum of the interior angles of any regular polygon of n sides is equal to 180(n - 2) degrees.
40 sides
It is a rhombus that has 4 equal sides but not 4 equal interior angles
It is an equilateral triangle that has 3 equal sides and 3 equal interior angle of 60 degrees.
An equilateral triangle has 3 equal sides and 3 equal interior angles (of 60o.)
if the interior angle is 150 the exterior angle equals 30. since exterior angle equals 360/ by number of sides the number of sides equals 360/30 which equal 12
The sum of the interior angles of any regular polygon of n sides is equal to 180(n - 2) degrees.
40 sides
How many sides does a regular polygon have if the measure of each interior angle is equal to 172 degress ?
It is a rhombus that has 4 equal sides but not 4 equal interior angles
It is an equilateral triangle that has 3 equal sides and 3 equal interior angle of 60 degrees.
An equilateral triangle has 3 equal sides and 3 equal interior angles (of 60o.)
Yes, an isosceles triangle can have an obtuse interior angle. An isosceles triangle has at least two equal sides and can have one angle greater than 90 degrees, making it obtuse. In such a case, the two equal angles would be acute, ensuring that the sum of the interior angles still equals 180 degrees.
A shape has three sides if its interior angles add up to equal 180 degrees.
The total interior angles of a 17-gon accumulate to 2700 degrees [180*(sides-2)]. Assuming all sides are equal, the total interior angle would be 2700/17 = 158.82
Mathematically - A regular pentagon has interior angles and sides equal. Bisecting each interior angle and extending a line segment to the center gives you five equal angles (360o/5=72o). Using the triangle formed by two line segements and a side of the pentagon we know the sum of interior angles of each triangle equals (180o) and we know one angle is 72o. The sum the other two interior angles of each triangle equals (180o-72o=108o) which equals each interior angle of the pentagon (equal to the sum of these two angles). Thus the exterior angle of each corner of the pentagon equals (360o-108o=252o). :-)
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